Problem: Simplify and expand the following expression: $ \dfrac{3}{3n - 6}+ \dfrac{2}{4n - 24}- \dfrac{n}{n^2 - 8n + 12} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{3}{3n - 6} = \dfrac{3}{3(n - 2)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{2}{4n - 24} = \dfrac{2}{4(n - 6)}$ We can factor the quadratic in the third term: $ \dfrac{n}{n^2 - 8n + 12} = \dfrac{n}{(n - 2)(n - 6)}$ Now we have: $ \dfrac{3}{3(n - 2)}+ \dfrac{2}{4(n - 6)}- \dfrac{n}{(n - 2)(n - 6)} $ The least common multiple of the denominators is: $ 12(n - 2)(n - 6)$ In order to get the first term over $12(n - 2)(n - 6)$ , multiply by $\dfrac{4(n - 6)}{4(n - 6)}$ $ \dfrac{3}{3(n - 2)} \times \dfrac{4(n - 6)}{4(n - 6)} = \dfrac{12(n - 6)}{12(n - 2)(n - 6)} $ In order to get the second term over $12(n - 2)(n - 6)$ , multiply by $\dfrac{3(n - 2)}{3(n - 2)}$ $ \dfrac{2}{4(n - 6)} \times \dfrac{3(n - 2)}{3(n - 2)} = \dfrac{6(n - 2)}{12(n - 2)(n - 6)} $ In order to get the third term over $12(n - 2)(n - 6)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{n}{(n - 2)(n - 6)} \times \dfrac{12}{12} = \dfrac{12n}{12(n - 2)(n - 6)} $ Now we have: $ \dfrac{12(n - 6)}{12(n - 2)(n - 6)} + \dfrac{6(n - 2)}{12(n - 2)(n - 6)} - \dfrac{12n}{12(n - 2)(n - 6)} $ $ = \dfrac{ 12(n - 6) + 6(n - 2) - 12n} {12(n - 2)(n - 6)} $ Expand: $ = \dfrac{12n - 72 + 6n - 12 - 12n}{12n^2 - 96n + 144} $ $ = \dfrac{6n - 84}{12n^2 - 96n + 144}$ Simplify: $ = \dfrac{n - 14}{2n^2 - 16n + 24}$